\documentclass[twoside]{article} \usepackage{amsfonts} % font used for R in Real numbers \usepackage{graphicx} % for including a postscript file \pagestyle{myheadings} \markboth{\hfil Solution to a semilinear problem \hfil EJDE--2003/08} {EJDE--2003/08\hfil Petr Tomiczek \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2003}(2003), No. 08, pp. 1--10. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Solution to a semilinear problem on type II regions determined by the Fucik spectrum % \thanks{ {\em Mathematics Subject Classifications:} 35J70, 58E05, 49B27. \hfil\break\indent {\em Key words:} Resonance, eigenvalue, jumping nonlinearities, Fucik spectrum. \hfil\break\indent \copyright 2003 Southwest Texas State University. \hfil\break\indent Submitted October 7, 2002. Published January 28, 2003. \hfil\break\indent Supported by Ministry of Education of the Czech Republic, MSM 235200001. } } \date{} % \author{Petr Tomiczek} \maketitle \begin{abstract} We prove the existence of solutions to the semi-linear problem $$\displaylines{ u''(x)+\alpha u^+(x)+\beta u^-(x)=f(x)\,,\quad x\in (0,\pi)\,,\cr u(0)=u(\pi)=0 }$$ where the point $(\alpha,\beta)$ falls in regions of type (II) between curves of the Fu\v{c}\'{\i}k spectrum. We use a variational method based on the generalization of the Saddle Point Theorem. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \section{Introduction} We investigate the existence of solutions for the nonlinear boundary-value problem \begin{eqnarray} \label{1.1} &u''(x)+ \alpha u^+(x)-\beta u^-(x)=f(x) \,,\quad x\in (0,\pi)\,,&\\ &x(0)=x(\pi)=0\,.& \nonumber \end{eqnarray} Here $u^{\pm}=\max\{\pm u,0\}$, $\alpha,\beta \in \mathbb{R}$ and $f\in L^2(0,\pi)$. For $f\equiv 0$, $\alpha=\lambda_+$, and $\beta=\lambda_-$ Problem (\ref{1.1}) becomes \begin{eqnarray}\label{1.2} &u''(x)+ \lambda_+ u^+(x)-\lambda_- u^-(x)=0\,,\quad x\in (0,\pi)\,,&\\ &x(0)=x(\pi)=0\,.& \nonumber \end{eqnarray} We define $\Sigma=\{(\lambda_+,\lambda_-)\in\mathbb{R}^2 : \mbox{(\ref{1.2}) has a nontrivial solution}\}$. This set is called the Fu\v{c}\'{\i}k spectrum (see \cite{lit3}), and can be expressed as $\Sigma=\bigcup_{j=1}^{\infty}\Sigma_j$ where \begin{eqnarray} \Sigma_1 & = &\{(\lambda_+ ,\lambda_- ) \in \mathbb{R}^2: (\lambda_+ -1)(\lambda_- -1) = 0 \}\,, \nonumber\\ \Sigma_{2i} & = & \{(\lambda_+,\lambda_-) \in \mathbb{R}^2: i \Bigl({1\over {\sqrt {\lambda_+}}}+{1\over {\sqrt {\lambda_-}}} \Bigr) = 1 \}\,, \nonumber\\ \Sigma_{2i+1} & = & \Sigma_{2i+1,1} \cup \Sigma_{2i+1,2} \qquad \mbox{where}\nonumber \\ \Sigma_{2i+1,1} & = & \{(\lambda_+,\lambda_-) \in \mathbb{R}^2: i\Bigl({1\over {\sqrt {\lambda_+}}}+ {1\over {\sqrt {\lambda_-}}} \Bigr)+ {1\over {\sqrt {\lambda_+}}}= 1\} \,,\nonumber\\ \Sigma_{2i+1,2} & = & \{(\lambda_+,\lambda_-) \in \mathbb{R}^2: i\Bigl({1\over {\sqrt {\lambda_+}}}+ {1\over {\sqrt {\lambda_-}}}\Bigr)+ {1\over {\sqrt {\lambda_-}}}= 1 \} \,.\nonumber \end{eqnarray} Note that there are two types of regions between the curves of $\Sigma$: \begin{description} \item [Type (I)] ${\cal R}_1$ which consists of regions between the curves $\Sigma_{2i}$ and $\Sigma_{2i+1}$, $i\in\mathbb{N}$. \item[Type (II)] ${\cal R}_2$ which consists of regions between the curves $\Sigma_{2i+1,1}$ and $\Sigma_{2i+1,2}$, $i\in\mathbb{N}$. \end{description} If $(\alpha,\beta)\in {\cal R}_1$ one can solve (\ref{1.1}) for arbitrary $f\in L^2(0,\pi)$ while this is not so for regions ${\cal R}_2$ (see \cite{lit1}). We suppose that the point $(\alpha,\beta)\in{\cal R}_2$ is between the curves $\Sigma_{2i+1,1}$ and $\Sigma_{2i+1,2}$, $\alpha>\beta$, and $k>0$ such that $\lambda_+=\alpha+k$, $\lambda_-=\beta+k$, $(\lambda_+,\lambda_-)\in\Sigma_{2i+1,2}$ and $\lambda_+<(2i+2)^2$. We denote $\varphi_2$ the solution of (\ref{1.2}) belonging to $(\lambda_+,\lambda_-)$. In this work we obtain existence results for (\ref{1.1}) with right hand side $f=-c\varphi_2+\varphi^\bot_2$, $c>0$ where $\int_0^{\pi} \varphi_2 \varphi^\bot_2 \,dx =0$ and $\bigl(\frac{1}{\beta-(2i)^2}+\frac{1}{(2i+2)^2-\alpha)}\bigr) \int_0^{\pi}(\varphi_2^\bot)^2\,dx< \bigl(\frac{1}{\lambda_+-\alpha}-\frac{1}{(2i+2)^2-\alpha)}\bigr) \int_0^{\pi} (c\varphi_2)^2\,dx$. This article is inspired by a result in \cite{lit7} where the author solves the problem $\Delta u(x)+\alpha u^+(x)+\beta u^-(x)+p(x,u(x))=f(x)$ with nontrivial nonlinearity satisfying $p(x,u(x))\not= 0$. \begin{remark} \label{rem1} \rm Assuming that $(2i+2)^2>\lambda_+>\lambda_-$, if $(\lambda_+,\lambda_-)\in \Sigma_{2i+1,1}$, then $\lambda_->(2i)^2$. Assuming that $(2i+2)^2>\alpha$, if the point $(\alpha,\beta)$ is in ${\cal R}_2$ between the curves $\Sigma_{2i+1,1}$ and $\Sigma_{2i+1,2}$, then $\beta >(2i)^2$. See the illustration in figure 1. \end{remark} \begin{figure}[t] \begin{center} \includegraphics[width=0.7\textwidth]{fig1.eps} \end{center} \caption{Regions determined by the Fucik spectrum} \end{figure} \section{Preliminaries} \paragraph{Notation:} We shall use the classical spaces ${C}(0,\pi)$, $L^p(0,\pi)$ of continuous and measurable real-valued functions whose $p$-th power of the absolute value is Lebesgue integrable, respectively. $H$ is the Sobolev space of absolutely continuous functions $u\colon (0,\pi)\to \mathbb{R}$ such that $u'\in L^2(0,\pi)$ and $u(0)=u(\pi)=0$. We denote by the symbols $\| \cdot \|$, and $\| \cdot \|_2$ the norm in $H$, and in $L^2(0,\pi)$, respectively. We denote $\langle ., . \rangle$ the pairing in the space $H$. By a solution of (\ref{1.1}) we mean a function $u\in {C}^1(0,\pi)$ such that $u'$ is absolutely continuous, $u$ satisfies the boundary conditions and the equations (\ref{1.1}) holds a.e. in $(0,\pi)$. Let $I:H\to \mathbb{R}$ be a functional such that $I\in {C}^1(H,\mathbb{R})$ (continuously differentiable). We say that $u$ is a critical point of $I$, if $$\langle I'(u), v\rangle = 0 \quad \mbox{for all}\quad v\in H\,.$$ We say that $\gamma$ is a critical value of $I$, if there is $u_0\in H$ such that $I(u_0)=\gamma$ and $I'(u_0)=0$. We say that $I$ satisfies Palais-Smale condition (PS) if every sequence $(u_n)$ for which $I(u_n)$ is bounded in $H$ and $I'(u_n)\to 0$ (as $n\to \infty)$ possesses a convergent subsequence. We study (\ref{1.1}) by using of varitional methods. More precisely, we look for critical points of the functional $I:H\to \mathbb{R}$, which is defined by $$\label{2.1} I(u)=\frac{1}{2}\int_0^{\pi }\bigl[(u')^2-\alpha(u^+)^2- \beta(u^-)^2\bigr] \,dx + \int_0^{\pi } fu \,dx\,.$$ Every critical point $u\in H$ of the functional $I$ satisfies $$\langle I'(u),v \rangle=\int_0^{\pi}\bigl[ u'v'-(\alpha u^+- \beta u^-)v +fv\bigr] \,dx=0\qquad \mbox{for all}\quad v\in H\,.$$ Then $u$ is also a weak solution of (\ref{1.1}) and vice versa. The usual regularity argument for ODE yields immediately (see Fu\v{c}\'{\i}k \cite{lit3}) that any weak solution of (\ref{1.1}) is also the solution in the sense mentioned above. \paragraph{Method:} We will use the following variant of Saddle Point Theorem which is proved in Struwe \cite[Theorem 8.4]{lit6}. \begin{theorem} \label{thm1} Let $S$ be a closed subset in $H$ and $Q$ a bounded subset in $H$ with relative boundary $\partial Q$. Set $\Gamma = \{h : h\in {\bf C}(H,H), h(x)=x\mbox{ on } \partial Q\}$. Suppose $I\in {C}^1(H,\mathbb{R})$ and \begin{description} \item[$(i)$] $S \cap \partial Q = \emptyset$, \item[$(ii)$] $S \cap h(Q) \not= \emptyset$, for every $h\in\Gamma$, \item[$(iii)$] there are constants $\mu,\nu$ such that $\mu=\inf_{u\in S} I(u) > \sup_{u\in \partial Q} I(u)=\nu$, \item[$(iv)$] $I$ satisfies Palais-Smale condition. \end{description} Then the number $$\gamma=\inf_{h\in \Gamma}\sup_{u\in Q} I(h(u))$$ defines a critical value $\gamma>\nu$ of $I$. \end{theorem} \paragraph{Remark:} We say that $S$ and $\partial Q$ {\it link} if they satisfy conditions i), ii) of the theorem above. Now we present a few results needed later. \begin{lemma} \label{l1} Let $\varphi$ be a solution of (\ref{1.2}) with $(\lambda_+,\lambda_-) \in \Sigma$, $\lambda_+\ge\lambda_-$ and $u=a\varphi+w$, $a\ge0$, $w\in H$. Then functional $J(u)= \int_0^\pi \bigl[(u')^2-\lambda_+(u^+)^2- \lambda_-(u^-)^2\bigr] \,dx$ satisfies $$\label{2.2} \int_0^{\pi }\bigl[(w')^2-\lambda_+w^2\bigr]\,dx \le J(u)\le \int_0^{\pi }\bigl[(w')^2-\lambda_-w^2 \bigr] \,dx\,.$$ \end{lemma} \paragraph{Proof:} We prove the inequality in the right of (\ref{2.2}), the proof of the inequality in the left is similar. Since $\varphi$ is the solution of (\ref{1.2}) it holds $$\label{2.3} \int_0^\pi (\varphi')^2 \, dx=\int_0^\pi \bigl[\lambda_+(\varphi^+)^2+\lambda_-(\varphi^-)^2\bigr] \,dx$$ and $$\label{2.4} \int_0^\pi \varphi'w' \, dx=\int_0^\pi \bigl[\lambda_+ \varphi^+ w -\lambda_-\varphi^- w \bigr] \,dx\qquad \mbox{for}\quad w\in H\,.$$ By (\ref{2.3}) and (\ref{2.4}) we obtain \begin{eqnarray} \label{2.5} J(u) &=& \int_0^{\pi} \Bigl[((a\varphi+w)')^2 -\lambda_+ ((a\varphi+w)^+)^2-\lambda_-((a\varphi+w)^-)^2\Bigr]\, dx\nonumber \\ &= & \int_0^{\pi} \Bigl[(a\varphi')^2 +2a\varphi'w'+(w')^2- (\lambda_+-\lambda_-)((a\varphi+w)^+)^2 \nonumber \\ &&\quad -\lambda_-(a\varphi+w)^2\Bigr]\, dx \nonumber\\ &= &\int_0^{\pi} \Bigl[(\lambda_+-\lambda_-)(a\varphi^+)^2 +\lambda_-(a\varphi)^2+2a((\lambda_+-\lambda_-)\varphi^++\lambda_-\varphi)w \nonumber \\ &&\quad +(w')^2 -(\lambda_+-\lambda_-)((a\varphi+w)^+)^2-\lambda_-((a\varphi)^2 +2a\varphi w+w^2)\Bigr]\, dx \nonumber\\ &=& \int_0^{\pi} \Bigl\{(\lambda_+-\lambda_-)\bigl[(a\varphi^+)^2 +2a\varphi^+w-((a\varphi+w)^+)^2\bigr] \nonumber \\ && \quad +(w')^2-\lambda_-w^2\Bigr\}\, dx\,. \end{eqnarray} For the function $(a\varphi^+)^2+2a\varphi^+w-((a\varphi+w)^+)^2$ in the last integral in (\ref{2.5}) it holds \begin{eqnarray*} \lefteqn{(a\varphi^+)^2+2a\varphi^+w-((a\varphi+w)^+)^2}\\ &=&\left\{ \begin{array}{ll} -((a\varphi+w)^+)^2\le 0 & \varphi<0 \\ -w^2\le 0 & \varphi\ge0\,, a\varphi+w\ge 0 \\ a\varphi^+(a\varphi^++w+w)\le 0 & \varphi\ge0\,, a\varphi+w< 0\,. \\ \end{array}\right. \end{eqnarray*} Hence and by assumption $\lambda_+\ge\lambda_-$ we obtain the assertion of the lemma (\ref{l1}). \hfill$\square$ \begin{lemma} \label{l2} Let $\varphi$ be a solution of (\ref{1.2}) with $(\lambda_+,\lambda_-)\in\Sigma$ and $\varphi^\bot\in H$ such that $\int_0^{\pi} \varphi \varphi^\bot \,dx =0$. Let $d>0$ and $w\in H$ satisfying $\int_0^{\pi} [(w')^2-\lambda_- w^2]\,dx \le -d \int_0^{\pi} w^2 \,dx$. We put $u=a\varphi+w$, $a\ge0$ and $I(u)=\frac{1}{2}\int_0^{\pi} [(u')^2- (\lambda_+-k)(u^+)^2-(\lambda_- -k)(u^-)^2 -2(c\varphi+\varphi^\bot) u ]\, dx$ where $0< k \le d$ and $c>0$. Then there is constant $\tilde{a}>0$ such that for $u=\tilde{a}\varphi+w$ it holds \begin{eqnarray}\label{2.6} I(u)&\le& -\frac{1}{2}\int_0^{\pi} \bigl[ \frac{1}{k}(c\varphi)^2-\frac{1}{d-k}(\varphi^\bot)^2 \bigr]\, dx \end{eqnarray} \end{lemma} \paragraph{Proof:} By (\ref{2.2}) from Lemma \ref{l1} and the assumptions on $w$, we obtain \begin{eqnarray}\label{2.7} \lefteqn{I(u)}\nonumber\\ &=& \frac{1}{2}\int_0^{\pi} \bigl[(u')^2- \lambda_+(u^+)^2- \lambda_- (u^-)^2\bigr]\, dx+ \int_0^{\pi} \bigl[ \frac{k}{2}u^2-(c\varphi+\varphi^\bot) u\bigr]\, dx \nonumber \\ &\le &\frac{1}{2}\int_0^{\pi} \bigl[(w')^2- \lambda_-w^2\bigr]\, dx + \int_0^{\pi} \bigl[\frac{k}{2}u^2-(c\varphi+\varphi^\bot) u\bigr]\, dx \le \nonumber\\ &\le& \int_0^{\pi} \bigl[- \frac{d}{2} w^2+\frac{k}{2}u^2-(c\varphi+\varphi^\bot) u\bigr]\, dx\,. \end{eqnarray} We substitute $u=a\varphi+w$ and we have for the last integral in (\ref{2.7}) \begin{eqnarray} \lefteqn{\int_0^{\pi} \bigl[ -\frac{d}{2}w^2+\frac{k}{2}u^2-(c\varphi+\varphi^\bot) u \bigr]\, dx } \nonumber \\ &=& \int_0^{\pi} \bigl[ -\frac{d}{2}w^2+\frac{k}{2}(a\varphi+w)^2 -(c\varphi+\varphi^\bot)(a\varphi+w) \bigr]\, dx \nonumber \\ &=& \int_0^{\pi} \bigl[ -\frac{1}{2}(d-k)w^2+a\big(\frac{k}{2}a-c\big)\varphi^2+(ak-c)\varphi w-\varphi^\bot w \bigr]\, dx\,.\nonumber \end{eqnarray} We put $a=\frac{c}{k} (=\tilde{a})$ and we obtain \begin{eqnarray}\label{2.8} \lefteqn{\int_0^{\pi}\bigl[ -\frac{1}{2}(d-k)w^2+\frac{c}{k} \big(\frac{k}{2}\frac{c}{k} -c\big)\varphi^2-\varphi^\bot w \bigr]\, dx }\nonumber \\ &=& -\frac{1}{2} \int_0^{\pi}\bigl[ \bigl(\sqrt{(d-k)}w+\frac{1}{\sqrt{(d-k)}}\varphi^\bot\bigl)^2 +\frac{1}{k}(c\varphi)^2-\frac{1}{d-k}(\varphi^\bot)^2\bigr]\, dx \le \nonumber \\ &\le& -\frac{1}{2} \int_0^{\pi}\bigl[ \frac{1}{k}(c\varphi)^2-\frac{1}{d-k}(\varphi^\bot)^2\bigr]\, dx\,. \end{eqnarray} From the above inequality and (\ref{2.7}) it follows the assertion of Lemma \ref{l2}. \hfill$\square$ \begin{lemma} \label{l3} Let $\widehat{\varphi}$ be a solution of (\ref{1.2}) with $(\widehat{\lambda}_+,\widehat{\lambda}_-)\in\Sigma$. Let $u=a\widehat{\varphi}+w$, $a\ge0$, $f\in L_2(0,\pi)$ and $w\in H$ such that $\int_0^{\pi} [(w')^2-\widehat{\lambda}_- w^2]\,dx \le 0$. Let $l>0$ then $\forall k_2<0$ $\exists K$ such that for $u$ with $\|u\|_2\ge K$ it holds $$\label{2.9} I(u)= \frac{1}{2}\int_0^{\pi} \bigl[(u')^2- (\widehat{\lambda}_++l)(u^+)^2- (\widehat{\lambda}_- +l)(u^-)^2+2fu\bigr]\, dx\le k_2 <0\,.$$ \end{lemma} \paragraph{Proof:} From inequality (\ref{2.2}) in Lemma \ref{l1} and H\"{o}lder inequality we obtain \begin{eqnarray} I(u)&=& \frac{1}{2}\int_0^{\pi} \bigl[(u')^2- \widehat{\lambda}_+(u^+)^2-\widehat{\lambda}_- (u^-)^2\bigr] \, dx +\int_0^{\pi}\bigl[-\frac{l}{2}u^2+fu\bigr]\,dx\nonumber\\ &\le& \frac{1}{2}\int_0^{\pi} \bigl[(w')^2- \widehat{\lambda}_-w^2 \bigr]\, dx -\frac{l}{2}\|u\|^2_2 + \|f\|_2 \|u\|_2\,.\nonumber \end{eqnarray} By assumption, the integral in the above inequality is less than zero. Then it is easy to see that for sufficiently large $K$ and $\|u\|_2>K$, it holds $I(u)\le k_2<0$. \section{Main result} \begin{theorem} \label{thm2} Let $(\alpha,\beta)$ be a point in ${\cal R}_2$ between the curves $\Sigma_{2i+1,1}$ and $\Sigma_{2i+1,2}$,. Let $\alpha>\beta$ and $k>0$ be such that $\lambda_+=\alpha+k$, $\lambda_-=\beta+k$, and $(\lambda_+,\lambda_-)\in \Sigma_{2i+1,2}$. Assume that $\lambda_+<(2i+2)^2$. We denote by $\varphi_2$ the solution of (\ref{1.2}) with $(\lambda_+,\lambda_-)$ and $\varphi^\bot_2$ be the function satisfying $\int_0^{\pi} \varphi_2 \varphi^\bot_2 \,dx =0$. We put $f=-c\varphi_2+\varphi^\bot_2$, with $c>0$ and we assume that \begin{eqnarray*} \lefteqn{\bigl(\frac{1}{\beta-(2i)^2}+\frac{1}{(2i+2)^2-\alpha)}\bigr) \int_0^{\pi}(\varphi_2^\bot)^2\,dx}\\ &<&\bigl(\frac{1}{\lambda_+-\alpha}-\frac{1}{(2i+2)^2-\alpha)}\bigr) \int_0^{\pi} (c\varphi_2)^2\,dx\,. \end{eqnarray*} Then there exists a solution of (\ref{1.1}). \end{theorem} \paragraph{Proof:} We take $l>0$ such that $(\alpha-l,\beta-l)\in\Sigma_{2i+1,1}$ and we denote by $\varphi_1$ the nontrivial solution of (\ref{1.2}) with $(\alpha-l,\beta-l)$. Let $H^-$ be the subspace of $H$ spanned by function $\sin x, \sin 2x, \ldots, \sin 2ix$. We define $V\equiv V_1\cup V_2$ where \begin{eqnarray*} V_1&=&\{u\in H : u=a_1\varphi_1+w,\; 0\le a_1, \; w\in H^-\}, \\ V_2&=&\{u\in H : u=a_2\varphi_2+w,\; 0\le a_2, \; w\in H^-\}. \end{eqnarray*} Let $\tilde{a}>0$, $L>0$ then we define $Q\equiv Q_1\cup Q_2$ where \begin{eqnarray*} Q_1&=&\{u\in V_1 : 0\le a_1\le \tilde{a},\; \|w\|\le L\},\\ Q_2&=&\{u\in V_2 : 0\le a_2\le \tilde{a},\; \|w\|\le L\}. \end{eqnarray*} Let $S$ be the subspace of $H$ spanned by function $\sin (2i+2)x,\sin (2i+3)x, \ldots$\,. Next, we verify the assumptions of Theorem \ref{thm1}. We see that $S$ is the closed subset of $H$ and $Q$ is the bounded subset in $H$. \noindent\textbf{i)} For $u\in \partial Q_1$ we have $u=a_1\varphi_1+w$ with $a_1>0$ and $\langle a_1\varphi_1+w,\sin(2i+1)x\rangle=a_1\langle \varphi_1,\sin(2i+1)x\rangle>0$. Similarly for $u\in \partial Q_2$ we obtain $\langle u,\sin(2i+1)x\rangle<0$. For $z\in S$ it holds $\langle z,\sin(2i+1)x\rangle=0$. Hence it follows $\partial Q\cap S =\emptyset$. \noindent\textbf{ii)} The proof of the assumption $S\cap h(Q)\not=\emptyset\quad \forall h\in\Gamma$ is similar to the proof in \cite[example 8.2]{lit7}. We see that $H=S\oplus V$ and let $\pi\colon H\to V$ be the continuous projection of $H$ onto $V$. We have to show that $0\in \pi(h(Q))$. For $t\in[0,1]$, $u\in Q$ we define $$h_t=t \pi(h(u))+(1-t)u\,.$$ Function $h_t$ defines a homotopy of $h_0= id$ with $h_1=\pi\circ h$. Moreover, $h_t|\partial Q=\mathop{\rm id}$ for all $t\in[0,1]$. Hence the topological degree $\deg(h_t,Q,0)$ is well-defined and by homotopy invariance we have $$\deg(\pi\circ h,Q,0)=\deg(id,Q,0)=1\,.$$ Hence $0\in \pi(h(Q))$, as was to be shown. \noindent\textbf{iii)} First we find the infimum of functional $I$ on the set $S$. We have \begin{eqnarray} I(u)&=&\frac{1}{2}\int_0^{\pi }\bigl[(u')^2- \alpha(u^+)^2-\beta(u^-)^2+ 2fu \bigr] \,dx \nonumber\\ &=&\frac{1}{2}\int_0^{\pi }\bigl[(u')^2-\alpha u^2+ (\alpha-\beta)(u^-)^2-2f u\bigr] \,dx \nonumber\,. \end{eqnarray} For $u\in S$ it holds $\int_0^{\pi } (u')^2 \,dx \ge (2i+2)^2\int_0^{\pi } u^2 \,dx$. We denote $b=(2i+2)^2-\alpha$ (which is positive by assumption) and we obtain \begin{eqnarray} I(u)&\ge& \frac{1}{2} \Big\{\int_0^{\pi}\bigl[bu^2- 2fu \bigr]\,dx +\int_0^{\pi}\bigl[(\alpha-\beta)(u^-)^2 \bigr] \,dx \Big\} \nonumber \\ &=& \frac{1}{2} \Big\{\int_0^{\pi}\bigl(\sqrt{b}u- \frac{1}{\sqrt{b}} f\bigr)^2 \,dx -\frac{1}{b}\int_0^{\pi} f^2 \,dx +\int_0^{\pi}(\alpha-\beta)(u^-)^2 \,dx \Big\} \nonumber\,. \end{eqnarray} We note that $\alpha>\beta$. Hence and from previous inequality it follows $$\label{3.1} \inf_{u\in S}I(u)\ge -\frac{1}{2b} \int_0^{\pi}f^2 \,dx\,= -\frac{1}{2((2i+2)^2-\alpha)} \int_0^{\pi}(c\varphi_2)^2 +(\varphi_2^\bot)^2\,dx\,.$$ Second we estimate the value $I(u)$ for $u\in\partial Q$. For a function $w\in H_-$, we have $$\|w\|^2=\int_0^{\pi} (w')^2\, dx \le (2i)^2 \int_0^{\pi} w^2\, dx =(2i)^2 \|w\|_2^2\,.$$ For $u\in Q_2$ $(u=a_2\varphi_2+w)$ we put $d=\lambda_- - (2i)^2>0$ then the function $w$ satisfies the assumption $\int_0^{\pi} (w')^2-\lambda_- w^2\, dx \le -d \int_0^{\pi} w^2\, dx$ of lemma\,\ref{l2}. It follows from the Remark \ref{rem1} that $d-k=(\lambda_--(2i)^2)-(\lambda_- -\beta)=\beta-(2i)^2>0$. We can use the inequality (\ref{2.6}) from the lemma\,\ref{l2} and for $u=\frac{c}{k}\varphi_2+w$ we obtain \begin{eqnarray} \label{3.2} I(u)&\le& -\frac{1}{2} \int_0^{\pi}\Bigl[ \frac{1}{k}(c\varphi)^2-\frac{1}{d-k}(\varphi^\bot)^2\Bigr]\, dx \nonumber\\ &=& -\frac{1}{2}\int_0^{\pi}\Bigl[\frac{1}{\lambda_+-\alpha} (c\varphi)^2- \frac{1}{\beta-(2i)^2}(\varphi^\bot)^2\Bigr]\, dx\,. \end{eqnarray} For $u\in Q_1$ we put $\widehat{\lambda}_+=\alpha-l$, $\widehat{\lambda}_- =\beta-l$, $l>0$ and $(\widehat{\lambda}_+,\widehat{\lambda}_-)\in \Sigma_{2i+1,1}$. It follows from remark\,\ref{rem1} that $\widehat{\lambda}_-\ge (2i)^2$ then the function $u\in\partial Q_1$ $(u=a_1\varphi_1+w)$ satisfies assumptions of lemma \ref{l3} with $f=-c\varphi_2+\varphi_2^\bot$. We put in (\ref{2.9}) $k_2=- \frac{1}{2}\int_0^{\pi}\Bigl[\frac{1}{\lambda_+-\alpha} (c\varphi)^2-\frac{1}{\beta-(2i)^2}(\varphi^\bot)^2\Bigr]\,dx$ and we obtain that the inequality (\ref{3.2}) holds for $u\in\partial Q_1$ too. Hence $$\label{3.3} \sup_{u\in\partial Q} I(u)\le - \frac{1}{2}\int_0^{\pi}\Bigl[\frac{1}{\lambda_+-\alpha} (c\varphi)^2-\frac{1}{\beta-(2i)^2}(\varphi^\bot)^2\Bigr]\,dx\,.$$ By (\ref{3.1}), (\ref{3.3}) and from the assumption of theorem \ref{thm2} we have \begin{eqnarray*} \inf_{u\in S} I(u)&\ge& -\frac{1}{2((2i+2)^2-\alpha)} \int_0^{\pi}(c\varphi_2)^2 +(\varphi_2^\bot)^2\,dx\\ &>&-\frac{1}{2} \int_0^{\pi}\Bigl[\frac{1}{\lambda_+-\alpha} (c\varphi)^2-\frac{1}{\beta-(2i)^2}(\varphi^\bot)^2\Bigr]\,dx \ge \sup_{u\in\partial Q} I(u)\,. \end{eqnarray*} Then assumption iii) of theorem \ref{thm2} holds. \noindent \textbf{iv)} For this assumption, we will show that $I$ satisfies the Palais-Smale condition. First, we suppose that the sequence $(u_n)$ is unbounded and there exists a constant $c$ such that $$\label{3.4} \Bigl| {1\over 2}\int_0^{\pi} \bigl[(u'_n)^2-\alpha (u_n^+)^2-\beta (u_n^-)^2 \bigr] \,dx +\int_0^{\pi} f u_n \,dx \Bigr| \le c$$ and $$\label{3.5} \lim_{n\to \infty}\| I'(u_n)\|=0\,.$$ Let $(w_k)$ be an arbitrary sequence bounded in $H$. It follows from (\ref{3.5}) and the Schwarz inequality that \begin{eqnarray} \label{3.6} \lefteqn{ \bigl| \lim_{{n\to \infty}\atop{k\to \infty}} \int_0^{\pi}\bigl[u_n'w_k'-(\alpha u_n^+ -\beta u_n^-)w_k\bigr] \,dx +\int_0^{\pi} fw_k \,dx \bigr| }\nonumber \\ &=&|\lim_{{n\to \infty}\atop{k\to \infty}}\langle I'(u_n), w_k\rangle | \le \lim_{{n\to \infty}\atop{k\to \infty}} \| I'(u_n)\|\cdot\| w_k\| =0\,. \end{eqnarray} Put $v_n=u_n/ \| u_n \|$. Due to compact imbedding $H\subset L^2(0,\pi)$ there is $v_0\in H$ such that (up to subsequence) $v_n\rightharpoonup v_0$ weakly in H, $v_n\to v_0$ strongly in $L^2(0,\pi)$. We divide (\ref{3.6}) by $\|u_n\|$ and we obtain $$\label{3.7} \lim_{{n\to \infty}\atop{k\to \infty}} \int_0^{\pi}\bigl[ v_n'w_k'-(\alpha v_n^+ -\beta v_n^-)w_k \bigr] \,dx=0$$ and also $$\label{3.8} \lim_{{n\to \infty}\atop{{m\to \infty}\atop{k\to \infty}}} \int_0^{\pi}\bigl[ (v_n'-v_m')w_k'-[\alpha (v_n^+-v_m^+) -\beta (v_n^--v_m^-)]w_k \bigr] \,dx=0\,.$$ We set $w_k=v_n-v_m$ in (\ref{3.8}) and we get $$\label{3.9} \lim_{{n\to \infty}\atop{m\to \infty}} \Bigl[\| v_n-v_m\|^2-\int_0^{\pi}\bigl[ [\alpha (v_n^+-v_m^+) -\beta (v_n^--v_m^-)](v_n-v_m) \bigr] \,dx\Bigr]=0\,.$$ Since $v_n\to v_0$ strongly in $L^2(0,\pi)$ the integral in (\ref{3.9}) convergent to 0 and then $v_n$ is a Cauchy sequence in $H$ and $v_n\to v_0$ strongly in $H$ and $\|v_0\|=1$. It follows from (\ref{3.7}) and the usual regularity argument for ordinary differential equations (see Fu\v{c}\'{\i}k \cite{lit3}) that $v_0$ is the solution of the equation $v_0''+\alpha v_0^+ -\beta v_0^-=0$. From the assumption $(\alpha,\beta)\not\in \Sigma$ it follows that $v_0=0$. This is contradiction to $\|v_0\|=1$. This implies that the sequence $(u_n)$ is bounded. Then there exists $u_0\in H$ such that $u_n\rightharpoonup u_0$ in $H$, $u_n\to u_0$ in $L^2(0,\pi)$ (up to subsequence). It follows from the equality (\ref{3.6}) that $$\label{3.10} \lim_{{{n\to \infty}\atop{m\to \infty}}\atop{k\to \infty}} \int_0^{\pi}\bigl[(u_n-u_m)'w_k'- [\alpha (u_n^+-u_m^+) -\beta (u_n^--u_m^-)]w_k\bigr] \,dx =0\,.$$ We put $w_k=u_n-u_m$ in (\ref{3.10}) and the strong convergence $u_n \to u_0$ in $L^2(0,\pi)$ and (\ref{3.10}) imply the strong convergence $u_n \to u_0$ in $H$. This shows that the functional $I$ satisfies Palais-Smale condition and the proof of Theorem\,2 is complete. \begin{thebibliography}{00} \frenchspacing \bibitem{lit1} E. N. Dancer: {\em On the Dirichlet problem for weakly non-linear elliptic partial differential equations}, Proc. Roy. Soc. Edin. {\bf 76A} (1977), 283-300. \bibitem{lit2} D. G. de Figueiredo, W. M. Ni: {\em Pertubations of second order linear elliptic problems by nonlinearities without Landesman-Lazer conditions}, Nonlinear analysis, Theory, Methods \& Applications, Vol 3, No.5, 629--634, (1979). \bibitem{lit3} S. Fu\v{c}\'{\i}k: {\em Solvability of Nonlinear Equations and Boundary Value problems}, D.Reidel Publ. Company, Holland 1980. \bibitem{lit4} E. M. Landesman, A. C. Lazer: {\em Nonlinear perturbations of elliptic boundary value problems at resonance}, J. Math. Mech. 19(1970), 609--623. \bibitem{lit5} P. Rabinowitz: {\em Minmax methods in critical point theory with applications to differential equations}, CBMS Reg. Conf. Ser. in Math. no 65, Amer. Math. Soc. Providence, RI., (1986). \bibitem{lit6} M. Struwe: {\em Variational Methods}, Springer, Berlin, (1996). \bibitem{lit7} M. Schechter: {\em Type (II) regions between curves of the Fucik spectrum}, Nonlinear differ. equ. appl. 4(1997), 459-476. \end{thebibliography} \noindent {\sc Petr Tomiczek} \\ Department of Mathematics, University of West Bohemia,\\ Univerzitn\'{\i} 22, 306 14 Plze\v{n}, Czech Republic \\ e-mail: tomiczek@kma.zcu.cz \end{document}